Query On Using Mod in MQuery

Question

Query On Using Mod in MQuery

Closed this issue 3 years ago · 4 comments

Hi Heckmann

I have been sourcing for information on how to do the following in mongo but always fail till I saw your git.
Data (data format: id, price, flag)
id_1, 11, false
id_2, 22, false
id_3, 33, false,
id_4. 44. false
id_5, 55. false.
id_6, 66. false

I am using TS and will like even row be flag as true. How can I do it using mquery. Can help? Thank

Answer 1 · 2017-10-02T04:08:43.000Z

Can you please clarify your question? I don't understand.

Answer 2 · 2017-10-02T07:47:44.000Z

Hi Valeri Is easy to create a row_number column on MySQL but not on Mongo. I need to write a NoSQL query to update Flag to true whenever there is an EVEN row. Is it doable? I tried MongoBooster, it refuse to convert. Thanks for your attention. [image: Inline images 1]

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On 2 October 2017 at 12:08, Valeri Karpov ***@***.***> wrote: Can you please clarify your question? I don't understand. — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub <#90 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AC_OthtzrZxmF6zPxPivhT6AdMIKGUITks5soGHLgaJpZM4PqK4K> .

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Answer 3 · 2018-07-31T05:43:16.000Z

@ybchoo If I understand correctly, what you want is to update flag to true when price is even (mod 2 equals to 0)

You can do this:

mquery().where('price').mod([2,0]).update({$set: { flag: true }})
  .setOptions({ multi: true }).exec(callback)

you can use all of the following syntax

where('price').mod([2,0])
where('price').mod(2, 0)
mod('price', [2, 0])
mod('price', 2, 0)

Answer 4 · 2018-07-31T05:51:29.000Z

Thanks.

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On Tue, 31 Jul 2018 at 13:43, Fonger ***@***.***> wrote: @ybchoo <https://github.com/ybchoo> If I understand correctly, what you want is to update flag to true when price is even (mod 2 equals to 0) You can do this: mquery().where('price').mod([2,0]).update({$set: { flag: true }}) .setOptions({ multi: true }).exec(callback) — You are receiving this because you were mentioned. Reply to this email directly, view it on GitHub <#90 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AC_OtpjAMm2V27c1yl7RVUiUWCARXJuFks5uL-51gaJpZM4PqK4K> .

-- Rgds *_______________________________* CYB Important: This email is confidential and may be privileged. If you are not the intended recipient, please delete it and notify me immediately. Please DO NOT copy or use it for any purpose, or disclose its' contents to ANY other person. Thank you.